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Homework Help: Setting up a triple integral using cylindrical & spherical coordinates

  1. Jan 28, 2010 #1
    1. The problem statement, all variables and given/known data

    Inside the sphere x2 + y2 + z2 = R2 and between the planes z = [tex]\frac{R}{2}[/tex] and z = R. Show in cylindrical and spherical coordinates.

    2. Relevant equations

    [tex]\iiint\limits_Gr\,dz\,dr\,d\theta[/tex]

    [tex]\iiint\limits_G\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]

    3. The attempt at a solution

    [tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_?^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
    Is my upper limit for r correct? How do I find the lower limit for z?

    [tex]\int_0^{2\pi}\int_0^{\frac{\pi}{3}}\int_?^R\rho^{2}sin\,\theta\,d\rho\,d\phi\,d\theta[/tex]
    How do I find the lower limit for rho?
     
  2. jcsd
  3. Jan 28, 2010 #2

    LCKurtz

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    Yes the upper limit is OK. Isn't the equation of the plane z = R/2? That would be z on the lower surface.

    Again, the equation of that plane is z = R/2. Express that in spherical coordinates and solve it for [itex]\rho[/itex]. You will find that [itex]\rho[/itex] depends on [itex]\phi[/itex], but that's OK.
     
  4. Jan 31, 2010 #3
    Ah, I didn't understand that concept. Thank you!

    [tex]2\int_0^{2\pi}\int_0^{R\sqrt{\frac{3}{4}}}\int_\frac{R}{2}^{\sqrt{R^{2}-r^{2}}}r\,dz\,dr\,d\theta[/tex]
    Is there supposed to be a 2 out in front? I don't know why I put it there and I don't think it is needed for this problem.
     
  5. Jan 31, 2010 #4

    LCKurtz

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    I don't know why you put it there either. Go with your instincts. If you don't think it should be there, don't put it there. :cool:
     
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